如何在多态字段中使用HaskellDB？(重叠实例的问题)

Question

我有一个有6种不同类型实体的模式,但它们都有很多共同点.我想我可能在类型级别抽象了很多这种共性,但我遇到了HaskellDB和重叠实例的问题.这是我开始使用的代码,工作正常:

import Database.HaskellDB
import Database.HaskellDB.DBLayout

data Revision a = Revision deriving Eq
data Book = Book

instance FieldTag (Revision a) where
  fieldName _ = "rev_id"

revIdField :: Attr (Revision Book) (Revision Book)
revIdField = mkAttr undefined

branch :: Table (RecCons (Revision Book) (Expr (Revision Book)) RecNil)
branch = baseTable "branch" $ hdbMakeEntry undefined
bookRevision :: Table (RecCons (Revision Book) (Expr (Revision Book)) RecNil)
bookRevision = baseTable "book_revision" $ hdbMakeEntry undefined

masterHead :: Query (Rel (RecCons (Revision Book) (Expr (Revision Book)) RecNil))
masterHead = do
  revisions <- table bookRevision
  branches <- table branch
  restrict $ revisions ! revIdField .==. branches ! revIdField
  return revisions

这工作正常,但branch太具体了.我实际想要表达的是以下内容:

branch :: Table (RecCons (Revision entity) (Expr (Revision entity)) RecNil)
branch = baseTable "branch" $ hdbMakeEntry undefined

但是,通过此更改,我收到以下错误:

Overlapping instances for HasField
                            (Revision Book)
                            (RecCons (Revision entity0) (Expr (Revision entity0)) RecNil)
  arising from a use of `!'
Matching instances:
  instance [overlap ok] HasField f r => HasField f (RecCons g a r)
    -- Defined in Database.HaskellDB.HDBRec
  instance [overlap ok] HasField f (RecCons f a r)
    -- Defined in Database.HaskellDB.HDBRec
(The choice depends on the instantiation of `entity0'
 To pick the first instance above, use -XIncoherentInstances
 when compiling the other instance declarations)
In the second argument of `(.==.)', namely `branches ! revIdField'
In the second argument of `($)', namely
  `revisions ! revIdField .==. branches ! revIdField'
In a stmt of a 'do' expression:
      restrict $ revisions ! revIdField .==. branches ! revIdField

我已经尝试盲目地投掷-XOverlappingInstances并且-XIncoherentInstances在此,但这没有帮助(我想真正理解为什么用类型变量替换具体类型会导致这个问题).

任何帮助和建议将不胜感激!

Answer 1

随着这个问题的出现，现在回答可能已经太晚了，无法对您产生任何影响，但也许如果其他人也遇到类似的问题......

归结为这样一个事实：无法推断出您想要entity引用Bookwhenbranch用于 中masterHead。错误消息中显示的部分

选择取决于“entity0”的实例化

告诉你哪里需要消除歧义，特别是你需要提供更多关于entity0应该是什么的信息。您可以提供一些类型注释来帮助解决问题。

首先，定义branch为

type BranchTable entity = Table (RecCons (Revision entity) (Expr (Revision entity)) RecNil)
branch :: BrancTable entity
branch = baseTable "branch" $ hdbMakeEntry undefined

然后改为masterHead阅读

masterHead :: Query (Rel (RecCons (Revision Book) (Expr (Revision Book)) RecNil))
masterHead = do
  revisions <- table bookRevision
  branches <- table (branch :: BranchTable Book)
  restrict $ revisions ! revIdField .==. branches ! revIdField
  return revisions

请注意应用于branch:的类型注释branch :: BranchTable Book，它用于消除导致类型错误的歧义。

要使其masterHead适用于Revision e其中包含字段的任何内容，您可以使用以下定义：

masterHead :: (ShowRecRow r, HasField (Revision e) r) => Table r -> e -> Query (Rel r)
masterHead revTable et =
  do  revisions <- table revTable
      branches <- table branch'
      restrict $ revisions ! revIdField' .==. branches ! revIdField'
      return revisions
  where (branch', revIdField') = revBundle revTable et
        revBundle :: HasField (Revision e) r => Table r -> e -> (BranchTable e, Attr (Revision e) (Revision e))
        revBundle table et = (branch, revIdField)

et需要参数来指定类型应该是什么，e并且可以归因undefined于正确的类型，如

masterHead bookRevision (undefined :: Book)

生成 SQL

SELECT rev_id1 as rev_id
FROM (SELECT rev_id as rev_id2
      FROM branch as T1) as T1,
     (SELECT rev_id as rev_id1
      FROM book_revision as T1) as T2
WHERE rev_id1 = rev_id2

虽然这确实需要FlexibleContexts，但它可以应用于提问者的模块，而无需重新编译 HaskellDB。